Wolfram Alpha symbolic integer derivative to this question is wrong.
How can i obtain it by Mathematica? Thanks
D[1/(-x + x*Exp[I*Pi*x]), {x, n}]
Wolfram Alpha symbolic integer derivative to this question is wrong.
How can i obtain it by Mathematica? Thanks
D[1/(-x + x*Exp[I*Pi*x]), {x, n}]
From Maple:
$\sum _{m=0}^n \binom{n}{m} (-m)_m x^{-1-m} \frac{\partial ^{m-n}}{\partial x^{m-n}}\frac{1}{\exp (i \pi x)-1}$
Finding n-th derivative:
$f(x)=\frac{1}{-1+e^{i \pi x}}$
f[x_] := 1/(Exp[Pi*I*x] - 1)
func = SeriesCoefficient[f[x], {x, 0, m}]
func = FullSimplify[func, {m \[Element] Integers, m > 0}]
$$f(x)=\sum _{m=-1}^{\infty } \text{SeriesCoefficient}[f(x),\{x,0,m\}] x^m$$ $$\frac{\partial ^nf(x)}{\partial x^n}=\sum _{m=-1}^{\infty } \text{SeriesCoefficient}[f(x),\{x,0,m\}] \frac{\partial ^nx^m}{\partial x^n}$$ $$\frac{\partial ^nf(x)}{\partial x^n}=\sum _{m=-1}^{\infty } \text{SeriesCoefficient}[f(x),\{x,0,m\}] \binom{m}{m-n} n! x^{m-n}$$
Sum[func*Binomial[m, m - n]*n!*x^(m - n), {m, -1, Infinity}]
MMA can't find sum :(. Changing index m
to m-1
in sum to be m=0
:
$$\frac{\partial ^n\frac{1}{(-1+e^{i \pi x})}}{\partial x^n}=\sum _{m=0}^{\infty } \frac{\left((i \pi )^{m-1} B_m\right) \binom{m-1}{m-n-1} n! x^{m-n-1}}{\Gamma (1+m)}$$
Substituting to first sum.
$$\frac{\partial ^n}{\partial x^n}(\frac{1}{-x+x \exp (i \pi x)})=\sum _{m=0}^n \binom{n}{m} (-m)_m x^{-1-m} \sum _{j=0}^{\infty } \frac{\left((i \pi )^{j-1} B_j\right) \binom{j-1}{j-n+m-1} (n-m)! x^{j-n+m-1}}{\Gamma (1+j)}$$
Check numerics:
n = 2;
N[D[1/(-x + x*Exp[I*Pi*x]), {x, n}] /. x -> 1, 50]
(*-1.0000000000000000000000000000000000000000000000000 -
1.5707963267948966192313216916397514420985846996876 I*)
N[Sum[Binomial[n, m]*Pochhammer[-m, m]*x^(-1 - m)*
Sum[((I \[Pi])^(j - 1) BernoulliB[j])/Gamma[1 + j]*
Binomial[j - 1, j - n + m - 1]*(n - m)!*x^(j - n + m - 1), {j,
0, 1000}], {m, 0, n}] /. x -> 1, 50]
(*-1.0000000000000000000000000000000000000000000000000 -
1.5707963267948966192313216916397514420985846996876 I*)
It seems that the result is correct.
D[1/(-x + x*Exp[I*Pi*x]), {x, n}]
now works. $\endgroup$